What is the slope of the line tangent to $f(x) = -2x^{2}-3x+7$ at $x = -3$ ?
The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x+h)^{2}-3(x+h)+7) - (-2x^{2}-3x+7)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x^{2}+2x h+h^{2})-3(x+h)+7) - (-2x^{2}-3x+7)}{h}$ $ = \lim_{h \to 0} \frac{-2x^{2}-4(x h)-2h^{2}-3x-3h+7+2x^{2}+3x-7}{h}$ $ = \lim_{h \to 0} \frac{-4(x h)-2h^{2}-3h}{h}$ $ = \lim_{h \to 0} -4x-2h-3$ $ = -4x-3$ $ = (-4)(-3)-3$ $ = 9$